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प्रश्न
Following table shows the amount of sugar production (in lakh tonnes) for the years 1931 to 1941:
| Year | Production | Year | Production |
| 1931 | 1 | 1937 | 8 |
| 1932 | 0 | 1938 | 6 |
| 1933 | 1 | 1939 | 5 |
| 1934 | 2 | 1940 | 1 |
| 1935 | 3 | 1941 | 4 |
| 1936 | 2 |
Complete the following activity to fit a trend line by method of least squares:
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उत्तर
Let yt be the trend line represented by the equation
yt = a + bt
Let u = `(t - "Midvalue")/h`
Midvalue = 1936, h = 1
∴ u = `(t - 1936)/1` = t – 1936
| Year (t) | yt | u | u2 | u.yt |
| 1931 | 1 | – 5 | 25 | – 5 |
| 1932 | 0 | – 4 | 16 | 0 |
| 1933 | 1 | – 3 | 09 | – 3 |
| 1934 | 2 | – 2 | 04 | – 4 |
| 1935 | 3 | – 1 | 01 | – 3 |
| 1936 | 2 | 0 | 00 | 0 |
| 1937 | 8 | 1 | 01 | 8 |
| 1938 | 6 | 2 | 04 | 12 |
| 1939 | 5 | 3 | 09 | 15 |
| 1940 | 1 | 4 | 16 | 04 |
| 1941 | 4 | 5 | 25 | 20 |
| `sumy_t` = 33 | `sumu` = 0 | `sumu^2` = 110 | 44 |
The equation of trend line becomes,
yt = a' + b'u .......(1)
The normal equations are
`sumy_t = na^' + b^'sumu` .......(2)
`sumu.y_t = u^'sumu + b^'sumu^2` ......(3)
From equation (2), we get
∴ Normal equations are
33 = 11a' + 0.b'
⇒ 11a' = 33
⇒ a' = 3
From equation (3), we get
44 = a'.0 + 110.b'
⇒ b' = `44/110` = 0.4
∴ b' = 0.4
∴ The equation of the trend line is given by yt = 3 + (0.4)u.
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| 1975 | 0 | - 4 | 16 | 0 |
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| 1977 | 3 | - 2 | 4 | - 6 |
| 1978 | 8 | - 1 | 1 | - 8 |
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| `sumx_t` =47 | `sumu`=0 | `sumu^2=60` | `square` |
The equation of trend line is xt= a' + b'u.
The normal equations are,
`sumx_t = na^' + b^' sumu` ...(1)
`sumux_t = a^'sumu + b^'sumu^2` ...(2)
Here, n = 9, `sumx_t = 47, sumu= 0, sumu^2 = 60`
By putting these values in normal equations, we get
47 = 9a' + b' (0) ...(3)
40 = a'(0) + b'(60) ...(4)
From equation (3), we get a' = `square`
From equation (4), we get b' = `square`
∴ the equation of trend line is xt = `square`
